Vtyjkyuk

Theorem. There exists a well ordered set $A$ having a largest element $Omega$ such that the section $S_Omega$ of $A$ by $Omega$ is uncountable but every other section of $A$ is countable.

Proof. Consider an uncountable well ordered set $B$, consider $C=1,2times B$ with the lexicographic order and let $Omega $ be the smallest element of $C$ for which $S_Omega=xin C$ is uncountable. Let $A=S_OmegacupOmega$
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I'm trying to show that $S_Omega$ has no largest element.

Assume the converse: there is $Gammain S_Omega$ such that $xle Gamma$ for all $xin S_Omega$. I need to use that $Gamma $ is a largest element. It's essential to consider $S_Gamma$ ($Gammain C$, and $S_Gamma$ is the section of $C$ by $Gamma$). Then $S_Gamma$ is countable. At the same time $S_Omega$ is not. But I don't see any contradiction here... I guess I need to invoke $Omega$ and prove something like $S_OmegacupOmega=S_Gamma$. But I don't see why this holds. $$S_Gamma=xin C: xle Gamma
\ S_OmegacupOmega=xin C: xle Omega$$
Do I need to use $xle Gamma$ for all $xin S_Omega$ somehow to prove the equality? I don't see how.

Note that $S_GammacupGamma=S_Omega$ holds. Indeed, let $xin S_GammacupGamma$. If $x=Gamma$, then $xin S_Omega$ by the definition of $Gamma$. If $xin S_Gamma$, then $xin S_Omega$ because $Gamma in S_Omega$ (and so if $xle Gamma$, then $x < Omega$). Thus $S_GammacupGammasubseteq S_Omega$. Conversely, let $xin S_Omega$. Since $Gamma$ is a largest element, $xle Gamma$, which means $xin S_GammacupGamma$. Thus $S_GammacupGamma=S_Omega$.

Now $S_Omega$ is uncountable, so $S_Gamma$ must be uncountable. However, for any $tin C$ with $tne Omega$ (in particular for $t=Gamma$; note that $Gammane Omega$ because $Gammain S_Omega$ and $Omeganotin S_Omega$), $S_t$ is countable. This is a contradiction.